The Chemistry of Acids and Bases Chapter 17 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of

The Chemistry of The Chemistry of Acids and BasesAcids and BasesThe Chemistry of The Chemistry of Acids and BasesAcids and Bases

Chapter 17Chapter 17Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to: Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida

22

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

Strong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/BasesStrong and Weak Acids/Bases

• Generally divide acids and bases into Generally divide acids and bases into STRONG or WEAK ones.STRONG or WEAK ones.

STRONG ACID:STRONG ACID: HNO HNO33(aq) + H(aq) + H22O(liq) --->O(liq) --->

HH33OO++(aq) + NO(aq) + NO33--(aq)(aq)

HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.

33



• Generally divide acids and bases into Generally divide acids and bases into STRONG or WEAK ones.STRONG or WEAK ones.

STRONG ACID:STRONG ACID: HNO HNO33(aq) + H(aq) + H22O(liq) --->O(liq) --->

HH33OO++(aq) + NO(aq) + NO33--(aq)(aq)

HNOHNO33 is about 100% dissociated in water. is about 100% dissociated in water.

44


H2O

H—Cl

Cl-

H3O+

H2O

H—Cl

Cl-

H3O+

HNO3, HCl, H2SO4 and HClO4 are among the only known strong acids.


55


• Weak acids are much less than 100% ionized Weak acids are much less than 100% ionized

in water.in water.

One of the best known is acetic acid = One of the best known is acetic acid = CHCH33COCO22H = HOAcH = HOAc

HOAc(aq) + HHOAc(aq) + H22O(liq) O(liq) OAcOAc--(aq) + H(aq) + H33OO++(aq)(aq)

OAcOAc-- = CH = CH33COCO22-- = acetate ion = acetate ion


66


• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.

NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)


77


• Strong Base:Strong Base: 100% dissociated in 100% dissociated in water.water.

NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)


Other common strong Other common strong bases include KOH and bases include KOH and Ca(OH)Ca(OH)22..

CaO (lime) + HCaO (lime) + H22O -->O -->

Ca(OH)Ca(OH)22 (slaked lime) (slaked lime)CaOCaO

88


• Weak base:Weak base: less than 100% ionized less than 100% ionized in waterin water

One of the best known weak bases is One of the best known weak bases is ammoniaammonia

NHNH33(aq) + H(aq) + H22O(liq) O(liq) NH NH44++(aq) + OH(aq) + OH--

(aq)(aq)


99


• Weak base:Weak base: less than 100% less than 100% ionized in waterionized in water

One of the best known weak bases is One of the best known weak bases is ammoniaammonia

NHNH33(aq) + H(aq) + H22O(liq) O(liq) NH NH44++(aq) + OH(aq) + OH--

(aq)(aq)


1010


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES• The most general theory for common The most general theory for common

aqueous acids and bases is the aqueous acids and bases is the BRØNSTED - LOWRY BRØNSTED - LOWRY theorytheory

• ACIDS DONATE HACIDS DONATE H++ IONS IONS

• BASES ACCEPT HBASES ACCEPT H++ IONS IONS

1111


The Brønsted definition means NHThe Brønsted definition means NH33 is a is a BASEBASE in water — and water is itself an in water — and water is itself an ACIDACID

BaseAcidAcidBaseNH4

+ + OH-NH3 + H2OBaseAcidAcidBase

NH4+ + OH-NH3 + H2O

ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

1212


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is a BASEBASE in water — and water is itself in water — and water is itself

an an ACIDACID

BaseAcidAcidBaseNH4


NH4+ + OH-NH3 + H2O

1313


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is a BASEBASE in water — and water is itself in water — and water is itself

an an ACIDACID

NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related — related

by the gain or loss of Hby the gain or loss of H++

BaseAcidAcidBaseNH4


NH4+ + OH-NH3 + H2O

1414


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESNHNH33 is a is a BASEBASE in water — and water is itself an in water — and water is itself an ACIDACID

NHNH3 3 / NH/ NH44++ is a is a conjugate pairconjugate pair — related by the — related by the

gain or loss of Hgain or loss of H++

Every acid has a conjugate base - and Every acid has a conjugate base - and vice-versa.vice-versa.

BaseAcidAcidBaseNH4


NH4+ + OH-NH3 + H2O

1515


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESA strong acid is 100% dissociated.A strong acid is 100% dissociated.

Therefore, a Therefore, a STRONG ACIDSTRONG ACID—a good H—a good H++ donor—must donor—must have a have a WEAK CONJUGATE BASEWEAK CONJUGATE BASE—a poor H—a poor H++ acceptor. acceptor.

HNOHNO33(aq) + H(aq) + H22O(liq) O(liq) H H33OO++(aq) + NO(aq) + NO33--(aq)(aq)

STRONG ASTRONG A basebase acid acid weak Bweak B

Notice that every A-B reaction has two acids Notice that every A-B reaction has two acids and two bases!and two bases!

1616


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIES

We know from experiment that HNOWe know from experiment that HNO33 is a is a strong acid.strong acid.

1.1. It is a stronger acid than HIt is a stronger acid than H33OO++

2.2. HH22O is a stronger base than NOO is a stronger base than NO33--

WEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2OWEAK BASE

ACID

STRONG ACID

BASEH3O+ + NO3

-HNO3 + H2O

1717


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESAcetic acid is only 0.42% ionized when [HOAc] = Acetic acid is only 0.42% ionized when [HOAc] =

1.0 M. It is a 1.0 M. It is a WEAK ACIDWEAK ACID

HOAc + HHOAc + H22O O H H33OO++ + OAc + OAc--

WEAK AWEAK A basebase acid acid STRONG BSTRONG B

Because [HBecause [H33OO++] is small, this must mean] is small, this must mean

1.1. HH33OO++ is a stronger acid than HOAc is a stronger acid than HOAc

2.2. OAcOAc-- is a stronger base than H is a stronger base than H22OO

1818


Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions• Now we can describe

reactions of acids with bases and the direction of such reaction.

• Consider the acid HF reacting with the base NH3.

• HF + NH3 --> NH4+ + F-

• Now we can describe reactions of acids with bases and the direction of such reaction.


• HF + NH3 --> NH4+ + F-

1919


Acid-Base ReactionsAcid-Base ReactionsAcid-Base ReactionsAcid-Base Reactions• Now we can describe

reactions of acids with bases and the direction of such reaction.


• HF + NH3 --> NH4+ + F-

• Now we can describe reactions of acids with bases and the direction of such reaction.


• HF + NH3 --> NH4+ + F-

2020


Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions

Predicting the Direction of Predicting the Direction of Acid-Base Reactions Acid-Base Reactions

Based on experiment, we can put acids and bases on a Based on experiment, we can put acids and bases on a chart.chart.

See Table 17.3 (page 794)See Table 17.3 (page 794)

ACIDSACIDS CONJUGATE BASESCONJUGATE BASES

STRONGSTRONG weakweak

weakweak STRONGSTRONG

This chart can be used to predict the direction of This chart can be used to predict the direction of reactions between any A-B pair.reactions between any A-B pair.

Reactions always go from the stronger A-B pair to the Reactions always go from the stronger A-B pair to the weaker A-B pair.weaker A-B pair.

2121


ACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESACID-BASE THEORIESPredicting the direction of an acid-base reaction.Predicting the direction of an acid-base reaction.

ACID1 BASE 2 ACID2 BASE1+ +

STRONG weak

Use Table 17.3Use Table 17.3Reactions always go from Reactions always go from the stronger A-B pair to the stronger A-B pair to the weaker A-B pair.the weaker A-B pair.

2222


MORE ABOUT WATERMORE ABOUT WATERHH22O can function as both an ACID and a BASE.O can function as both an ACID and a BASE.

In pure water there can be In pure water there can be AUTOIONIZATIONAUTOIONIZATION

2323


MORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATERMORE ABOUT WATER

KKww = [H = [H33OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

In a In a neutralneutral solution [Hsolution [H33OO++] = [OH] = [OH--]]

so Kso Kww = [H = [H33OO++]]22 = [OH = [OH--]]22

and so [Hand so [H33OO++] = [OH] = [OH--] = 1.00 x 10] = 1.00 x 10-7-7 M M

OH-

H3O+

OH-

H3O+

AutoionizationAutoionization

2424


Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of

pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].

SolutionSolution

2 H2 H22O(liq) O(liq) H H3OO++(aq) + OH(aq) + OH--(aq)(aq)

Le Chatelier predicts equilibrium shifts to Le Chatelier predicts equilibrium shifts to the ____________. the ____________.

[H[H3OO++] < 10] < 10-7-7 at equilibrium. at equilibrium.

Set up a concentration table.Set up a concentration table.

2525


Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of pure You add 0.0010 mol of NaOH to 1.0 L of pure

water. Calculate [Hwater. Calculate [H3OO++] and [OH] and [OH--].].

SolutionSolution


initialinitial 00 0.00100.0010

changechange +x+x +x+x

equilibequilib xx 0.0010 + x0.0010 + x

KKww = (x) (0.0010 + x) = (x) (0.0010 + x)

Because x << 0.0010 M, assume Because x << 0.0010 M, assume [OH[OH--] = 0.0010 M] = 0.0010 M

[H[H3OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M

2626


Calculating [HCalculating [H33OO++] & [OH] & [OH--]]Calculating [HCalculating [H33OO++] & [OH] & [OH--]]You add 0.0010 mol of NaOH to 1.0 L of You add 0.0010 mol of NaOH to 1.0 L of

pure water. Calculate [Hpure water. Calculate [H3OO++] and [OH] and [OH--].].

SolutionSolution


[H[H3OO++] = K] = Kww / 0.0010 = 1.0 x 10 / 0.0010 = 1.0 x 10-11-11 M M

This solution is This solution is BASICBASIC because because [H[H3OO++] < [OH] < [OH--]]

2727


[H[H33OO++], [OH], [OH--] and pH] and pH[H[H33OO++], [OH], [OH--] and pH] and pH

A common way to express acidity and basicity A common way to express acidity and basicity is with pHis with pH

pH = log (1/ [HpH = log (1/ [H33OO++]) ]) = =

- log [H - log [H33OO++]]

In a neutral solution, In a neutral solution, [H[H3OO++] = [OH] = [OH--] = ] =

1.00 x 101.00 x 10-7-7 at 25 at 25 ooCC

pH = -log (1.00 x 10pH = -log (1.00 x 10-7-7) ) = = - (-7) = 7- (-7) = 7

2828



What is the pH of the What is the pH of the 0.0010 M NaOH solution? 0.0010 M NaOH solution?

[H[H3OO++] = 1.0 x 10] = 1.0 x 10-11-11 M M

pH = - log (1.0 x 10pH = - log (1.0 x 10-11-11) = 11.00) = 11.00

General conclusion —General conclusion —

Basic solution Basic solution pH > 7pH > 7

Neutral Neutral pH = 7pH = 7

Acidic solutionAcidic solution pH < 7pH < 7

2929



If the pH of Coke is 3.12, it is ____________.If the pH of Coke is 3.12, it is ____________.

Because pH = - log [HBecause pH = - log [H3OO++] then] then

log [Hlog [H3OO++] = - pH] = - pH

Take antilog and getTake antilog and get

[H[H3OO++] = 10] = 10-pH-pH

[H[H3OO++] = 10] = 10-3.12-3.12 = =

7.6 x 107.6 x 10-4-4 M M

3030


Other pX ScalesOther pX ScalesOther pX ScalesOther pX Scales

In generalIn general pX = -log XpX = -log X

and so and so pOH = - log [OHpOH = - log [OH--]]

KKww = [H = [H3OO++] [OH] [OH--] = 1.00 x 10] = 1.00 x 10-14-14 at 25 at 25 ooCC

Take the log of both sidesTake the log of both sides

-log (10-log (10-14-14) = - log [H) = - log [H3OO++] + (-log [OH] + (-log [OH--])])

14 = pH + pOH14 = pH + pOH

3131


Equilibria Involving Equilibria Involving Weak Acids and BasesWeak Acids and Bases


Aspirin is a good example Aspirin is a good example of a weak acid, of a weak acid, KKaa = 3.2 x 10 = 3.2 x 10-4-4

3232




AcidAcid Conjugate BaseConjugate Base

acetic, CHacetic, CH33COCO22HH CHCH33COCO22--, acetate, acetate

ammonium, NHammonium, NH44++ NHNH33, ammonia, ammonia

bicarbonate, HCObicarbonate, HCO33-- COCO33

2-2-, carbonate, carbonate

A weak acid (or base) is one that ionizes A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).to a VERY small extent (< 5%).

3333




Consider acetic acidConsider acetic acid

HOAc + HHOAc + H22O O H H33OO++ + OAc + OAc--

AcidAcid Conj. Conj. basebase

3434




Consider acetic acidConsider acetic acid

HOAc + HHOAc + H22O HO H33OO++ + OAc + OAc--

AcidAcid Conj. base Conj. base

(K is designated K(K is designated Kaa for for ACIDACID))

Because [HBecause [H33OO++] and [OAc] and [OAc--] are SMALL, K] are SMALL, Kaa << 1. << 1.

Ka [H3O+][OAc- ]

[HOAc] 1.8 x 10-5Ka

[H3O+][OAc- ][HOAc]

1.8 x 10-5

3535




Values of KValues of Kaa for acid and K for acid and Kbb for bases are for bases are

found in found in TABLE 17.4TABLE 17.4 — page 799 — page 799

Notice the relation of TABLE 17.4 to the Notice the relation of TABLE 17.4 to the

table of relative acid/base strengths table of relative acid/base strengths

(Table 17.3).(Table 17.3).

3636


Equilibria Involving A Weak Acid

Equilibria Involving A Weak Acid

Determining the pH of Determining the pH of

an acetic acid an acetic acid

solution.solution.

See Screen 17.8.See Screen 17.8.

Determining the pH of Determining the pH of

an acetic acid an acetic acid

solution.solution.

See Screen 17.8.See Screen 17.8.

0.0001 M0.0001 M

0.003 M0.003 M

0.06 M0.06 M

2.0 M2.0 M

a pH metera pH meter

3737


Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak Acid

You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.

Step 1.Step 1. Define equilibrium concs. Define equilibrium concs.

[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial

changechange

equilibequilib

3838



You have 1.00 M HOAc. Calc. the You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hequilibrium concs. of HOAc, H33OO++, OAc, OAc--, ,

and the pH.and the pH.


[HOAc][HOAc] [H[H33OO++]] [OAc[OAc--]]

initialinitial 1.001.00 00 00

changechange -x-x +x+x +x+x

equilibequilib 1.00-x1.00-x xx xx

Note that we neglect [HNote that we neglect [H33OO++] from H] from H22O.O.

3939



Step 2.Step 2. Write K Write Kaa expression expression

You have 1.00 M HOAc. Calc. the equilibrium concs. You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, Hof HOAc, H33OO++, OAc, OAc--, and the pH., and the pH.

4040





Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

4141




This is a quadratic. Solve using quadratic formula or method of This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A).approximations (see Appendix A).


Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

4242



Step 3.Step 3. Solve K Solve Kaa expression expression


Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

4343




First assume x is very small because KFirst assume x is very small because Kaa is so small. is so small.


Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

4444






Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

4545





And so x = [And so x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2


Ka 1.8 x 10-5 = [H3O+][OAc- ]

[HOAc]

x2

1.00 - xKa 1.8 x 10-5 =

[H3O+][OAc- ][HOAc]

x2

1.00 - x

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

4646



Step 3.Step 3. Solve K Solve Kaa approximateapproximate expression expression

x = [x = [HH33OO++] = [] = [OAcOAc--] = [K] = [Kaa • 1.00] • 1.00]1/21/2


Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

4747





x = x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M


Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

4848





x = x = [[HH33OO++] = [] = [OAcOAc--] = 4.2 x 10] = 4.2 x 10-3-3 M M

pH = - log [pH = - log [HH33OO++] = -log (4.2 x 10] = -log (4.2 x 10-3-3) = ) = 2.372.37


Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00

4949



Consider the approximate expression Consider the approximate expression

x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2Ka 1.8 x 10-5 =

x2

1.00Ka 1.8 x 10-5 =

x2

1.00

5050




For many weak acidsFor many weak acids

[H[H33OO++] = [conj. base] = [K] = [conj. base] = [Kaa • C • Coo]]1/21/2

where Cwhere C00 = initial conc. of acid = initial conc. of acid

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00 x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2

5151







Useful Rule of Thumb:Useful Rule of Thumb:

x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2Ka 1.8 x 10-5 =

x2

1.00Ka 1.8 x 10-5 =

x2

1.00

5252







Useful Rule of Thumb:Useful Rule of Thumb:

If 100 • KIf 100 • Kaa < C < Coo, then [H, then [H33OO++] = [K] = [Kaa • C • Coo]]1/21/2

Ka 1.8 x 10-5 = x2

1.00Ka 1.8 x 10-5 =

x2

1.00 x [H3O+ ] = [Ka • 1.00]1/2x [H3O+ ] = [Ka • 1.00]1/2

5353


Equilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidEquilibria Involving A Weak AcidCalculate the pH of a 0.0010 M solution of Calculate the pH of a 0.0010 M solution of

formic acid, HCOformic acid, HCO22H.H.

HCOHCO22H + HH + H22O O HCO HCO22-- + H + H33OO++

KKaa = 1.8 x 10 = 1.8 x 10-4-4

Approximate solutionApproximate solution

[H[H33OO++] ] = = [K[Kaa • C • Coo]]1/21/2 = 4.2 x 10 = 4.2 x 10-4-4 M, M, pH = 3.37pH = 3.37

Exact SolutionExact Solution

[H[H33OO++] = [] = [HCOHCO22--] = 3.4 x 10] = 3.4 x 10-4-4 M M

[[HCOHCO22HH] = 0.0010 - 3.4 x 10] = 0.0010 - 3.4 x 10-4-4 = 0.0007 M = 0.0007 M

pH = 3.47 pH = 3.47

5454


Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak Base

You have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.

NHNH33 + H + H22O O NH NH44++ + OH + OH--

KKbb = 1.8 x 10 = 1.8 x 10-5-5


[NH[NH33]] [NH[NH44++]] [OH[OH--]]

initialinitial

changechange

equilibequilib

5555





KKbb = 1.8 x 10 = 1.8 x 10-5-5


[NH[NH33]] [NH[NH44++]] [OH[OH--]]



equilibequilib 0.010 - x0.010 - x x x xx

5656





KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 2.Step 2. Solve the equilibrium expression Solve the equilibrium expression

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

5757





KKbb = 1.8 x 10 = 1.8 x 10-5-5


Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so x x

= [OH= [OH--] = [NH] = [NH44++] = 4.2 x 10] = 4.2 x 10-4-4 M M

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

5858





KKbb = 1.8 x 10 = 1.8 x 10-5-5


Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so x = x =

[OH[OH--] = [NH] = [NH44++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 0.010 M 0.010 M

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

5959


Equilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseEquilibria Involving A Weak BaseYou have 0.010 M NHYou have 0.010 M NH33. Calc. the pH.. Calc. the pH.


KKbb = 1.8 x 10 = 1.8 x 10-5-5


Assume x is small (100•KAssume x is small (100•Kbb < C < Coo), so ), so x = [OHx = [OH--] ]

= [NH= [NH44++] = 4.2 x 10] = 4.2 x 10-4-4 M M

and [NHand [NH33] = 0.010 - 4.2 x 10] = 0.010 - 4.2 x 10-4-4 0.010 M 0.010 M

The approximation is validThe approximation is valid !!

Kb 1.8 x 10-5 = [NH4

+][OH- ][NH3 ]

= x2

0.010 - xKb 1.8 x 10-5 =

[NH4+][OH- ]

[NH3 ] =

x2

0.010 - x

6060





KKbb = 1.8 x 10 = 1.8 x 10-5-5

Step 3.Step 3. Calculate pH Calculate pH

[OH[OH--] = 4.2 x 10] = 4.2 x 10-4-4 M M

so pOH = - log [OHso pOH = - log [OH--] = 3.37] = 3.37

Because pH + pOH = 14,Because pH + pOH = 14,

pH = 10.63pH = 10.63

6161


MX + HMX + H22O ----> acidic or basic solution?O ----> acidic or basic solution?

Consider NHConsider NH44ClCl

NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(a)(a) Reaction of ClReaction of Cl-- with H with H22OO

ClCl-- + + HH22O ---->O ----> HCl + HCl + OHOH--

basebase acidacid acidacid basebase

ClCl-- ion is a VERY weak base because its ion is a VERY weak base because its conjugate acid is strong. conjugate acid is strong.

Therefore, ClTherefore, Cl-- ----> neutral solution ----> neutral solution

Acid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of SaltsAcid-Base Properties of Salts

6262


NHNH44Cl(aq) ----> NHCl(aq) ----> NH44++(aq) + Cl(aq) + Cl--(aq)(aq)

(b)(b) Reaction of NHReaction of NH44++ with H with H22OO

NHNH44++ + H + H22O ---->O ----> NHNH33 + + HH33OO++

acidacid basebase basebase acidacid

NHNH44++ ion is a moderate acid because its ion is a moderate acid because its

conjugate base is weak. conjugate base is weak.

Therefore, NHTherefore, NH44++ ----> acidic solution ----> acidic solution

See TABLE 17.5 for a summary of See TABLE 17.5 for a summary of acid-base properties of ions.acid-base properties of ions.


6363


Calculate the pH of a 0.10 M solution of NaCalculate the pH of a 0.10 M solution of Na22COCO33. .

NaNa++ + H + H22O ---> neutralO ---> neutral

COCO332-2- ++ HH22OO HCOHCO33

-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 1.Step 1. Set up concentration tableSet up concentration table

[CO[CO332-2-]] [HCO[HCO33

--]] [OH[OH--]]

initialinitial

changechange

equilibequilib


6464





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 1.Step 1. Set up concentration tableSet up concentration table

[CO[CO332-2-]] [HCO[HCO33

--]] [OH[OH--]]



equilibequilib 0.10 - x0.10 - x xx xx


6565





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 2.Step 2. Solve the equilibrium expressionSolve the equilibrium expression


6666





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4



Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x

6767





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo


Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x

6868





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


Assume 0.10 - x 0.10, because 100•KAssume 0.10 - x 0.10, because 100•Kbb < C < Coo

x = [HCOx = [HCO33--] = [OH] = [OH--] = 0.0046 M ] = 0.0046 M


Kb = 2.1 x 10-4 = [HCO3

- ][OH- ]

[CO32 ]

x2

0.10 - xKb = 2.1 x 10-4 =

[HCO3- ][OH- ]

[CO32 ]

x2

0.10 - x

6969





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4

Step 3.Step 3. Calculate the pHCalculate the pH

[OH[OH--] = 0.0046 M] = 0.0046 M


7070





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


[OH[OH--] = 0.0046 M] = 0.0046 M

pOH = - log [OHpOH = - log [OH--] = 2.34] = 2.34


7171





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


[OH[OH--] = 0.0046 M] = 0.0046 M


pH + pOH = 14, pH + pOH = 14,


7272





-- ++ OHOH--


KKbb = 2.1 x 10 = 2.1 x 10-4-4


[OH[OH--] = 0.0046 M] = 0.0046 M


pH + pOH = 14, pH + pOH = 14,

so so pH = 11.66pH = 11.66, and the solution is ________., and the solution is ________.


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The Chemistry of Acids and Bases Chapter 17 Copyright © 1999 by Harcourt Brace & Company All rights reserved. Requests for permission to make copies of